Practice Questions

Q81.Let f(x) = 5 −|x −2| and g(x) = |x + 1|, x ∈ R. If f(x) attains maximum value at α and g(x) attains (x−1)(x2−5x+6) minimum value at β, then lim is equal to x→−αβ x2−6x+8 (1) 3 (2) 1 2 2 (3) −32 (4) −12

Q81.A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is tan−1( 21 ). Water is poured into it at a constant rate of 5 cubic m/min. Then the rate (in m/min), at which the level of water is rising at the instant when the depth of water in the tank is 10 m; is: (1) 1 (2) 1 10π 15π (3) 1 (4) 2 5π π

Q81.The maximum volume in 𝑐𝑢. 𝑚 of the right circular cone having slant height 3 𝑚 is: JEE Main 2019 (09 Jan Shift 1) JEE Main Previous Year Paper (1) 2√3 𝜋 (2) 3√3 𝜋 4 (3) 6 𝜋 (4) 3𝜋

Q81.If the function f given by f(x) = x3 −3(a −2)x2 + 3ax + 7, for some a ∈R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, f(x)−14 = 0, (x ≠1) is : (x−1)2 (1) 7 (2) −7 (3) 6 (4) 5

Q81.If 𝑆1 and 𝑆2 are respectively the sets of local minimum and local maximum points of the function, 𝑓𝑥= 9𝑥4 + 12𝑥3 - 36𝑥2 + 25, 𝑥∈𝑅, then (1) 𝑆1 = -2; 𝑆2 = {0,1} (2) 𝑆1 = -1; 𝑆2 = 0,2 (3) 𝑆1 = -2,0; 𝑆2 = {1} (4) 𝑆1 = -2,1; 𝑆2 = {0}

Q81.For x > 1, if (2x)2y = 4e2x−2y , then (1 + loge 2x)2 dxdy is equal to (1) loge2x (2) xloge2x−loge2x (3) xloge2x (4) xloge2x+loge2x

Q81.If 𝑓1 = 1, 𝑓'1 = 3, then the derivative of 𝑓𝑓𝑓𝑥+ 𝑓𝑥2 at 𝑥= 1 is: JEE Main 2019 (08 Apr Shift 2) JEE Main Previous Year Paper (1) 9 (2) 12 (3) 15 (4) 33

Q82.Let S be the set of all values of x for which the tangent to the curve y = f(x) = x3 −x2 −2x at (x, y) is parallel to the line segment joining the points (1, f(1)) and (−1, f(−1)), then S is equal to (1) {−13 , −1} (2) {−13 , 1} (3) { 31 , 1} (4) { 13 , −1} 3 xdx is equal to Q83. ∫sec2x ⋅cot 4 3 x + C (1) 3tan−13 x + C (2) −34 tan−4 (3) −3tan−13 x + C (4) −3cot−13 x + C π/2 sin3x dx is:

Q82.The maximum area (in sq. units) of a rectangle having its base on the x− axis and its other two vertices on the parabola, y = 12 −x2 such that the rectangle lies inside the parabola, is : (1) 20√2 (2) 32 (3) 36 (4) 18√3

Q82.A spherical iron ball of radius 10 𝑐𝑚 is coated with a layer of ice of uniform thickness that melts at a rate of 50 𝑐𝑚3 / 𝑚𝑖𝑛. When the thickness of the ice is 5 𝑐𝑚, then the rate at which the thickness ( in 𝑐𝑚/ 𝑚𝑖𝑛) of the ice decreases, is : 1 1 (1) (2) 9π 36π (3) 1 (4) 5 18π 6π

Q82.If ∫x5e−4x3dx = 481 e−4x3f(x) + C , where C is a constant of integration, then f(x) is equal to (1) −4x3 −1 (2) −2x3 + 1 (3) −2x3 −1 (4) 4x3 + 1 π/2 dx where [t] denotes the greatest integer less than or equal to t, is

Q82.If 𝜃 denotes the acute angle between the curves, 𝑦= 10 - 𝑥2 and 𝑦= 2 + 𝑥2 at a point of their intersection, then tan⁡𝜃 is equal to: (1) 4 (2) 8 9 17 7 8 (3) (4) 17 15

Q82.Let f be a differentiable function from R to R such that |f(x) −f(y)| ≤2|x −y|3/2, for all x, y ∈R. If 1 f(0) = 1 then ∫ f 2(x)dx is equal to 0 (1) 0 (2) 1 (3) 2 (4) 21

Q82.The integral ∫ 3x13+2x11 dx, is equal to (2x4+3x2+1)4 (1) x4 + C (2) x4 + C 6(2x4+3x2+1)3 (2x4+3x2+1)3 (3) x12 + C (4) x12 + C (2x4+3x2+1)3 6(2x4+3x2+1)3 e x e x dx is equal to

Q82.The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is: 2 (1) √3 (2) 3√3 (3) √6 (4) 2 √3

Q82.The shortest distance between the point ( 23 , 0) and the curve y = √x, (x > 0) , is (1) √3 (2) 5 2 4 (3) 3 (4) √5 2 2 π

Q82.If ∫ x+1 dx = f(x)√2x −1 + C, where C is a constant of integration, then f(x) is equal to: √2x−1 (1) 3 1 (x + 1) (2) 32 (x + 2) (3) 3 2 (x −4) (4) 31 (x + 4)

Q82.If ∫ dx = A(tan−1( x−13 ) + x2−2x+10f(x) ) (x2−2x+10)2 (1) A = 271 and f(x) = 9(x −1) (2) A = 811 and f(x) = 3(x −1) (3) A = 541 and f(x) = 9(x −1)2 (4) A = 541 and f(x) = 3(x −1) JEE Main 2019 (10 Apr Shift 1) JEE Main Previous Year Paper

Q82.If ∫esecx(secx tan xf(x) + (secx tan x + sec2x))dx = esecxf(x) + C, then a possible choice of f(x) is: (1) secx −tanx −12 (2) secx + tanx + 12 (3) xsecx + tanx + 12 (4) secx + xtanx −12

Q82.Let α ∈(0, π2 ) , be constant.If the integral ∫ tanx−tantanx+tanαα dx = A(x)cos2α + B(x)sin2α + C , where C is a constant of integration, then the functions A(x) and B(x) are respectively (1) x −α and loge|sin(x −α)| (2) x + α and loge|cos(x −α)| (3) x + α and loge|sin(x + α)| (4) x −α and loge|cos(x −α)| JEE Main 2019 (12 Apr Shift 2) JEE Main Previous Year Paper α+1 dx 9 = loge( 8 ) is

Q82.The integral ∫2𝑥3 - 1 is equal to 𝑥4 + 𝑥𝑑𝑥, (1) 2 (2) |𝑥3 + 1| 1 (𝑥3 + 1) + 𝐶 + 𝐶 log𝑒 𝑥2 2log𝑒 |𝑥3| (3) 𝑥3 + 1 (4) 1 |𝑥3 + 1| log𝑒 𝑥 + 𝐶 2log𝑒 𝑥2 + 𝐶

Q82.Themaximum value of the finction f(x) = 3x3 −18x2 + 27x −40 on the set S = {x ∈R : x2 + 30 ≤11x} is : (1) -122 (2) -222 (3) 122 (4) 222 JEE Main 2019 (11 Jan Shift 1) JEE Main Previous Year Paper + C, for a suitable chosen integer m and a function A(x), where C is a

Q83.Given that the slope of the tangent to a curve 𝑦= 𝑦( 𝑥) at any point 𝑥, 𝑦 is 2𝑦𝑥2. If the curve passes through the centre of the circle 𝑥2 + 𝑦2 - 2𝑥- 2𝑦= 0, then its equation is (1) 𝑥2log𝑒⁡|𝑦| = - 2(𝑥- 1) (2) 𝑥log𝑒⁡|𝑦| = 2(𝑥- 1) (3) 𝑥log𝑒⁡|𝑦| = - 2(𝑥- 1) (4) 𝑥log𝑒⁡|𝑦| = 𝑥- 1 1

Q83.If x = 3 tant and y = 3 sect, then the value of dx2d2y π at t = 4 , is: (1) 1 (2) 1 6 6√2 (3) 1 (4) 3 3√2 2√2

Q83.If ∫𝑥5𝑒-𝑥2𝑑𝑥= 𝑔𝑥𝑒-𝑥2 + 𝑐, where 𝑐 is a constant of integration, then 𝑔-1 is equal to 5 (1) - (2) -1 2 (3) 1 (4) -1 2 π 2 4 3